Question

Assume that the mass has been moving along its circular path for some time. You start timing its motion with a stopwatch when it crosses the positive x axis, an instant that corresponds to t=0. [Notice that when t=0, r⃗ (t=0)=Ri^.] For the remainder of this problem, assume that the time t is measured from the moment you start timing the motion. Then the time − t refers to the moment a time t before you start your stopwatch.
What is the velocity of the mass at a time − t?
Express this velocity in terms of R, ω, t, and the unit vectors i^ and j^.

Answer 1

Answer:

$v = R\omega(-sin\omega t \hat i + cos\omega t \hat j)$

Explanation:

As we know that the mass is revolving with constant angular speed in the circle of radius R

So we will have

$\theta = \omega t$

now the position vector at a given time is

$r = Rcos\theta \hat i + R sin\theta \hat j$

now the linear velocity is given as

$v = \frac{dr}{dt}$

$v = (-R sin\theta \hat i + R cos\theta \hat j)\frac{d\theta}{dt}$

$v = R\omega(-sin\omega t \hat i + cos\omega t \hat j)$